Problem: Simplify the following expression: $y = \dfrac{-3x^2+4x- 1}{-3x + 1}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-3)}{(-1)} &=& 3 \\ {a} + {b} &=& &=& {4} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $3$ and add them together. The factors that add up to ${4}$ will be your ${a}$ and ${b}$ When ${a}$ is ${1}$ and ${b}$ is ${3}$ $ \begin{eqnarray} {ab} &=& ({1})({3}) &=& 3 \\ {a} + {b} &=& {1} + {3} &=& 4 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-3}x^2 +{1}x) + ({3}x {-1}) $ Factor out the common factors: $ x(-3x + 1) - 1(-3x + 1)$ Now factor out $(-3x + 1)$ $ (-3x + 1)(x - 1)$ The original expression can therefore be written: $ \dfrac{(-3x + 1)(x - 1)}{-3x + 1}$ We are dividing by $-3x + 1$ , so $-3x + 1 \neq 0$ Therefore, $x \neq \frac{1}{3}$ This leaves us with $x - 1; x \neq \frac{1}{3}$.